Electric field due to given potentials is 

${\mathrm{E}}_1=\frac{-\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{z}}=10\mathrm{ }\left|\mathrm{z}\right|$ for $\mid \mathrm{z}\mid \le 1 \mathrm{m}$. Here  $E_1$ is proportional to $z$, so charge distribution is continuous for $\left|\mathrm{z}\right|\le 1 \mathrm{m}$

${\mathrm{E}}_2=-\frac{\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{z}}=10$ for $\left|\mathrm{z}\right|\ge 1\mathrm{ }\mathrm{m}$. Here, $E_2$ is constant. 

So, using the Poisson’s equation ${\nabla }^{2}V=-\frac{\rho }{{\epsilon }_0}$, we get $\rho =0$.

$\therefore$ The source is an infinity large non-conducting thick plate.  

Hence, consider a cylindrical Gaussian surface:

Applying Gauss's law for $\mid \mathrm{z}\mid \le 1 \mathrm{m}$

$\mathrm{E}\times 2\mathrm{A}=\frac{\mathrm{\rho }\times 2\mathrm{z}\times \mathrm{A}}{{\mathrm{\epsilon }}_0}$

$\Rightarrow \left(10\mathrm{z}\right)\left(2\mathrm{A}\right)=\frac{\mathrm{\rho }\left(2\mathrm{Az}\right)}{{\mathrm{\epsilon }}_0}$

$\Rightarrow {\mathrm{\rho }}_0=10{\mathrm{\epsilon }}_0$ for $\left|\mathrm{z}\right|\le 1\mathrm{ }\mathrm{m}$.